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<h1 id="example-of-mna---diodes">Example of MNA - Diodes</h1>

<p>A diode is an example of a <em>nonlinear</em> component. When connected between two nodes A and B, we can use the <a href="https://en.wikipedia.org/wiki/Diode_modelling">Shockley diode model</a> to model the voltage-current relationship.</p>
<p align="center"><img src="images/example_circuit_mna_dio.svg" alt="Diode definition" width="100px"></p>

<p>$$i_D = I_{SS}\left(e^\frac{v_A-v_B}{\eta V_T}\right)$$</p>
<p>The diode current flows <em>out</em> of node A, and <em>into</em> node B, so we have contributions to <em>two</em> current equations.</p>
<p>$$\begin{aligned}
f_A(...,v_A,...,v_B,...) &amp;= +i_D = I_S\left(e^{\frac{v_A-v_B}{nV_T}}-1\right)\
f_B(...,v_A,...,v_B,...) &amp;= -i_D=-I_S\left(e^{\frac{v_A-v_B}{nV-T}}-1)\right)
\end{aligned}$$</p>
<p>From this we can calculate the contributions to the <em>Y-matrix</em>. Since the current equation contributions only depend on nodes A and B, we only get contributions to four elements.</p>
<p>$$\begin{aligned}
Y_{A,A} &amp;= \left.\frac{\partial f_A}{\partial v_A}\right|^{(k)}=\frac{I_S}{nV_T}e^{\frac{v_A^{(k)}-v_B^{(k)}}{nV_T}} &amp;= &amp;+g_D\
Y_{A,B} &amp;= \left.\frac{\partial f_A}{\partial v_B}\right|^{(k)}=-\frac{I_S}{nV_T}e^{\frac{v_A^{(k)}-v_B^{(k)}}{nV_T}} &amp;= &amp;-g_D\
Y_{B,A} &amp;= \left.\frac{\partial f_B}{\partial v_A}\right|^{(k)}=-\frac{I_S}{nV_T}e^{\frac{v_A^{(k)}-v_B^{(k)}}{nV_T}} &amp;= &amp;-g_D\
Y_{B,B} &amp;= \left.\frac{\partial f_B}{\partial v_B}\right|^{(k)}=\frac{I_S}{nV_T}e^{\frac{v_A^{(k)}-v_B^{(k)}}{nV_T}} &amp;= &amp;+g_D
\end{aligned}$$</p>
<p>We then calculate the contributions to the <em>RHS-vector</em>:</p>
<p>$$\begin{aligned}
RHS_A &amp;= f_A(...,v_A^{(k)},...,v_B^{(k)},...)-\pmb J_A\pmb x^{(k)}\
 &amp;= I_S\left(e^\frac{v_A^{(k)}-v_B^{(k)}}{nV_T}-1\right)-g_D\cdot (v_A^{(k)}-v_B^{(k)})\ &amp;= +c_D\ RHS_B &amp;= f_B(...,v_A^{(k)},...,v_B^{(k)},...)-\pmb J_B\pmb x^{(k)}\ &amp;= -\left(I_S\left(e^\frac{v_A^{(k)}-v_B^{(k)}}{nV_T}-1\right)-g_D\cdot (v_A^{(k)}-v_B^{(k)})\right)\ &amp;= -c_D
\end{aligned}$$</p>
<p>We note that this time the <em>RHS-vector</em> contributions are <em>not</em> 0 for the current equations. This is again typical for <em>nonlinear</em> components. The solution will need to be found in multiple iterations.</p>
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